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We have the constrained maximisation problem:

A perfectly competitive firm produces one output with two inputs, capital $(k)$ and labour $(l)$. The rental cost of capital is equal to $r >0$ and the wage rate is equal to $w>0$. The production function is $f(k, l)=(k + 1)^α l^{1-α}$ with $0 <α<1$. The firm wishes to minimise total costs while achieving a given level of production $y>0$.

Now for the sake of this course we construct this as a maximisation, problem I am particularly interested in the case where $k=0$ and $l>0$. While they do specifically consider this case they don't write the non-negativity constraint into the Lagrangian, and i believe this effects the analysis at the end. I have attached a screenshot of the "answer."

UPDATE: The specific equation they derive in the screenshot actually goes on to be very important in the following question. So thoughts are very much appreciated, thank!

$\mathcal{L}(k, l, \lambda, \mu_k, \mu_l) = -rk - wl + \lambda [(k + 1)^\alpha l^{1 - \alpha} - y] + \mu_k k + \mu_l l$`

  1. With respect to k:

$\frac{\partial \mathcal{L}}{\partial k} = -r + \lambda \alpha (k + 1)^{\alpha - 1} l^{1 - \alpha} + \mu_k$

  1. With respect to l:

$\frac{\partial \mathcal{L}}{\partial l} = -w + \lambda (1 - \alpha) (k + 1)^\alpha l^{-\alpha} + \mu_l$

  1. With respect to λ:

$\frac{\partial \mathcal{L}}{\partial \lambda} = (k + 1)^\alpha l^{1 - \alpha} - y$

Now, consider the specific case where $k = 0$ and $l > 0$:`

  1. $\frac{\partial \mathcal{L}}{\partial k}\bigg|_{k=0} = -r + \lambda \alpha l^{1 - \alpha} + \mu_k \le 0$

  2. $\frac{\partial \mathcal{L}}{\partial l}\bigg|_{k=0} = -w + \lambda (1 - \alpha) l^{-\alpha} = 0$

  3. $\frac{\partial \mathcal{L}}{\partial \lambda}\bigg|_{k=0} = l^{1 - \alpha} - y = 0$

  • $k^* = 0$
  • $l^* = y^{\frac{1}{1-\alpha}}$
  • $λ = \frac{w}{1-\alpha}y^{\frac{\alpha}{1-\alpha}}$
  1. They solve for for a condition on $y$ (see screenshot) consistent with $k = 0$ but their lack of an explicit use of a $\mu_k$ for our non-negativity on $k$ results in a different condition to what i would get, which would include $\mu_k$ . Why have they not explicitly used $\mu_k$ and $\mu_l$ as i have??
  2. I also get a weird condition on $\mu_k$ by just substituting in $l^*$ and $λ^*$ does it looks right?
  • $\mu_k \le r - \frac{\alpha wy^{\frac{1}{1-\alpha}}}{1-\alpha}$

Screen Shot from Answer showing condition on y

enter image description here

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  • $\begingroup$ The production function is not well-defined for $l<0$ which would imply a fractional (in the sense of non-integer) power of a negative quantity - see here. $\endgroup$ May 4, 2023 at 13:05
  • $\begingroup$ Your FOC's 2 and 3 are equivalent to those in the screen shot, so should lead to the same formula for $\lambda$. Possibly it's just a typo, but your formula should have $y^{\frac{\alpha}{1-\alpha}}$ not $y^{\frac{1}{1-\alpha}}$. The "weird" condition on $\mu_k$ will then become $\mu_k \leq r-\frac{\alpha wy^{\frac{1}{1-\alpha}}}{1-\alpha}$. $\endgroup$ May 4, 2023 at 13:24
  • $\begingroup$ Hi Adam, thanks again for your help. I understand why we don't worry about $l < 0$, but regarding $k = 0$ and the negativity constraint i have tried to elaborate why it's bothering me below. Thanks! $\endgroup$
    – CormJack
    May 6, 2023 at 14:34
  • $\begingroup$ I just edited out the typos I spotted $\endgroup$ May 6, 2023 at 16:46

2 Answers 2

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The only thing that changes is the $k$ first order condition.

The one you got with the non-negativity constraint Lagrange multiplier is

$-r + \lambda \alpha L^{1-\alpha} + \mu_k \leq 0$

Substracting $\mu_k$, and since $\mu_k \geq 0$,

$-r + \lambda \alpha L^{1-\alpha} \leq - \mu_k \leq 0$

Therefore,

$-r + \lambda \alpha L^{1-\alpha} \leq 0$.

This last inequality is the one used in the screenshot’s solution.

This is consistent with $k = 0$ because as long as this last inequality holds, there exists values of $\mu_k \geq 0$ such that your inequality holds.

You already found those values of $\mu_k$.

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  • $\begingroup$ Nicolas this looks like what i might be looking for! Which would be amazing, i'll comment again when i've properly reviewed, thanks! $\endgroup$
    – CormJack
    May 6, 2023 at 17:11
  • $\begingroup$ Hi Nicholas, i've thought about this some more. It's so helpful. I wish i could upvote it twice! It's so obvious now, but this little trick is really helpful. Would you agree implicitly this is what they are doing with the screenshot? As surely when you test the case for k = 0, they should be using a non-negativity constraint. So this is how they have got rid of it. Thanks so much! Any other clever tricks like this to do with KKT, or constraints you think i should know? $\endgroup$
    – CormJack
    May 8, 2023 at 11:00
  • $\begingroup$ One thing i'm uncertain about is why don't they state the specific condition on $u_k$ that i found? And don't we need to show that $ u_k \ge 0$ i'm not sure how to show this here? $\endgroup$
    – CormJack
    May 8, 2023 at 11:25
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    $\begingroup$ @CormJack Probably in the context of the problem, the value of $\mu_k$ isn’t relevant. I think when the non-negativity constraint is binding, $\mu_k$ is the marginal utility of $k$, as this would be analogous to $\lambda$ being the marginal utility of the budget/production constraint in consumer/firm theory aka shadow price. $\endgroup$ May 8, 2023 at 15:53
  • $\begingroup$ Hi @NicolasTorres, hope you're well! I hadn't noticed there's the comment from you before it's actually very helpful. Thank you! I I have a quick question regarding a minimisation problem and the non-negativity constraint. If you get a chance to look, that would be amazing: economics.stackexchange.com/questions/55496/… $\endgroup$
    – CormJack
    Apr 22 at 19:26
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I'm expanding on my question here as it's still causing trouble. It's a real pain so any additional thoughts are very appreciated!!

  • I believe that given their is a possibility $k = 0$ then the Lagrangian should have a non-negativity constraint on $k$ i.e. $k \ge 0 \implies -k \le 0$

  • This means that their Lagrangian is missing this constraint and the corresponding multiplier $λ_2$ i.e. $ + λ_2(k)$

  • They clearly think this is needed as they consider the case for $k = 0$ i.e. this constraint is binding.

  • In which case i do not understand why they never use $λ_2$? I also don't see why it's trivial, as including $λ_2$ seems to ruin the analysis at the end. See in red in the screenshot.

enter image description here

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